Consider the reaction and conditions shown below.* Cd(OH)_2(s) Cd^2+ + 2(OH^-) [

Question

Consider the reaction and conditions shown below.* Cd(OH)_2(s) Cd^2+ + 2(OH^-) [Cd(OH)_2(s)] = [Cd^2+] = 10^-4.00 M, pH = 6.50, mu = 0.11 M a) Use thermodynamic data from Appendix A in your textbook to determine the value of the equilibrium constant, K, at 25 degree C. b) If n were increased to 0.15 M, would the reaction be more likely to proceed to the right, to the left, or neither? c) Calculate the Gibbs energy of reaction (in kJ/mol) for the conditions specified. Do not assume that the solution is ideal. d) Is the reaction spontaneous in the forward direction, the reverse direction, or at equilibrium under the conditions specified? Write forward, reverse, or equilibrium, then briefly explain.

Explanation / Answer

For the given reaction,

a) Equilibrium constant K,

K = [Cd2+][OH-]^2

with,

pH = 6.50

pOH = 14 - 6.50 = 7.50

pOH = -log[OH-]

[OH-] = 3.16 x 10^-8 M

[Cd2+] = 1 x 10^-4 M

So,

K = (0.11)^3.(1 x 10^-4)(3.16 x 10^-8)^2

   = 1.33 x 10^-22

b) If ionic strength is increased to 0.15 M, the reaction more likely to proceed to the left. K value would increase.

c) Free energy dG

dG = -RTlnK

      = -8.314 x 298 ln(1.33 x 10^-22)

      = 124.80 kJ

d) The reaction is spontaneous in reverse direction as dG is +ve. Cd(OH)2 has minimal solubility is solution which is apparent from the solubility and K value.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---